I am trying to compute the boundary measure of the halfspace $\{x \geq 0 : x \in \mathbb{R}^2 \}$ under the standard two-dimensional Gaussian measure.
Boundary measure (after a change of variables) $$=\lim_{h \to 0} \frac{\int_{\theta = -\frac{\pi}{2}}^{\frac{\pi}{2}}\int_{r=0}^{h\sec \theta} \frac{1}{2\pi}e^{\frac{-r^2}{2}}rdrd\theta}{h}$$
$$ = \frac{1}{\pi} \lim_{h \to 0} \frac{\int_{0}^{\frac{\pi}{2}}\left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta}{h}. $$
How do we calculate this limit?
My crude attempt: We can split the integral into two parts -- one where $\cos \theta >> h$ and another where $\cos \theta$ is not $>> h$:
$$ \int_{0}^{\frac{\pi}{2}}\left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta$$
$$ = \int_{0}^{\arccos{h}} \left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta \; + \int_{\arccos{h}}^{\frac{\pi}{2}} \left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta $$
$$ \approx \int_{0}^{\arccos{h}} \frac{h^2}{2 \cos^2 \theta}d\theta \; + \int_{\arccos{h}}^{\frac{\pi}{2}} \left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta $$
$$ = \frac{h^2}{2}\tan(\arccos(h)) + \int_{\arccos{h}}^{\frac{\pi}{2}} \left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta .$$
Hence,
$$ \lim_{h \to 0} \frac{\int_{0}^{\frac{\pi}{2}}\left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta}{h}$$
$$ \approx \lim_{h \to 0} \frac{h}{2}\tan(\arccos(h)) + \lim_{h \to 0} \frac{ \int_{\arccos{h}}^{\frac{\pi}{2}} \left( 1 - e^{-\frac{h^2}{2 \cos^2 \theta}} \right)d\theta}{h} $$
I don't see how to proceed after this.